The moment of inertia of a uniform cylinder of length land radius R about its perpendicular bisector isI.  What is the ratio\frac{l}{R}such that the moment of inertia is minimum ?

 

  • Option 1)

    \sqrt{\frac{3}{2}}

  • Option 2)

    \frac{\sqrt{3}}{2}

  • Option 3)

    1

  • Option 4)

    \frac{3}{\sqrt{2}}

 

Answers (1)

As we learnt in

Moment of inertia of uniform rod of length(l) -

I= \frac{Ml^{2}}{12}

- wherein

About axis passing through its centre & perpendicular to its length.

 

 We know that solid cylinder about an axis which is perpendicular bisector.

I=\frac{mR^{2}}{4}+\frac{mR^{2}}{12}=\frac{m}{4}\left [ R^{2}+\frac{l^{2}}{3} \right ]

I=\frac{m}{4}\left [ \frac{\pi lR^{2}}{\pi l}+\frac{R^{2}}{3} \right ]=\frac{m}{4}\left [ \frac{v}{\pi l}+\frac{l^{2}}{3} \right ]

\frac{dx}{dl}=0    \Rightarrow\ \;\frac{m}{4}\left [ \frac{-v}{\pi l^{2}}+\frac{2l}{3} \right ]=0\ \;\Rightarrow\ \;\frac{-v}{\pi l^{2}}+\frac{2l}{3}=0

\frac{v}{\pi l^{2}}=\frac{2l}{3}\ \Rightarrow \; v=\frac{2\pi l^{3}}{3}

\pi R^{2}l=2\frac{\pi l^{3}}{3}\ \; \Rightarrow\ \; R^{2}=\frac{2l^{2}}{3}

\frac{l^{2}}{R^{2}}=\frac{3}{2}

\frac{l}{R}=\sqrt{}\frac{3}{2}


Option 1)

\sqrt{\frac{3}{2}}

This is the correct option.

Option 2)

\frac{\sqrt{3}}{2}

This is an incorrect option.

Option 3)

1

This is an incorrect option.

Option 4)

\frac{3}{\sqrt{2}}

This is an incorrect option.

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